Reduce Piles with Repeated Half Removals

### Problem Perform exactly `k` operations on the stone piles. In each operation, choose any pile and remove `floor(pileSize/2)` stones from its current contents. The same pile may be selected repeatedly. Return the minimum possible total number of stones left after all required operations. ```cpp class Solution { public: int minStoneSum(vector<int>& piles, int k); }; ``` ### Examples **Example 1:** ```text piles = [8,3], k = 2 Output: 5 Explanation: Reducing 8 to 4 and then to 2 leaves 2+3 stones. ``` **Example 2:** ```text piles = [1,1], k = 3 Output: 2 Explanation: Every required operation removes zero stones. ``` ### Constraints - `1 <= piles.length <= 10^5`. - `1 <= piles[i] <= 10^4`; `1 <= k <= 10^5`. ### Clarifying Questions Remaining pile size is ceil(oldSize/2). Exactly k operations are made even when further removals are zero. ```hint Inspect the structure How do the successive removal amounts from one pile change? ``` ### Reasoning checkpoints - Justify a globally minimum remaining total across all choices of repeated pile operations. - Respect rounding and the exact operation count, including piles from which no more stones can be removed. ### Optional extensions How can the upper bound on pile sizes help reduce the cost of selecting each operation?